Organic Chemistry
Organic Chemistry
8th Edition
ISBN: 9781305580350
Author: William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher: Cengage Learning
Question
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Chapter 17, Problem 17.21P

(a)

Interpretation Introduction

Interpretation:

Structural formula of a compound with the given formula upon oxidation by potassium dichromate in aqueous sulphuric acid gives carboxylic acid has to be drawn.

Concept-Introduction:

Oxidation:

Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation.  Oxidation state of atom ion or molecule will increase in this process.  In simple it is the addition of oxygen.

Example:

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  1

Here Fe2+ oxidized to Fe3+ by the removal of a single electron.

Oxidation of alcohols:

Oxidation of alcohol gives aldehydes ketones carboxylic acid Primary alocohols on oxidation yield carboxylic acid.  Secondary alcohol on oxidation gives ketone.  Depending upon the reaction conditions primary alcohols can be oxidized to aldehyde or carboxylic acid.  In the formation of carboxylic acid the alcohol is first oxidized to aldehyde which is then oxidized to form carboxylic acid.  Oxidation is not possible in tertiary alcohol because there is no reactive C-H bond present adjacent to the C-OH group so oxidation cannot be done without breaking the bond.  Common oxidizing agents are dichromate, collins reagent, dimethyl sulfoxide, potassium permanganate etc.

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  2

(b)

Interpretation Introduction

Interpretation:

Structural formula of a compound with the given formula upon oxidation by potassium dichromate in aqueous sulphuric acid gives carboxylic acid has to be drawn.

Concept-Introduction:

Oxidation:

Loss of electrons or loss of hydrogen from an atom ion or molecule during a chemical reaction is known as oxidation.  Oxidation state of atom ion or molecule will increase in this process.  In simple it is the gain of oxygen.

Example:

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  3

Here Cu2 oxidized to Cu2+ by the removal of two electrons.

Oxidation of alcohols:

Oxidation of alcohol gives aldehydes ketones carboxylic acid Primary alcohols on oxidation yield carboxylic acid.  Secondary alcohol on oxidation gives ketone.  If mild oxidizing agents are used the the reaction will end with an aldehydic product.  In the formation of carboxylic acid the alcohol is first oxidized to aldehyde which is then oxidized to form carboxylic acid.  Oxidation is not possible in tertiary alcohol because there is no reactive C-H bond present adjacent to the C-OH group so oxidation cannot be done without breaking the bond.  Common oxidizing agents are potassium dichromate, collins reagent, dimethyl sulfoxide, potassium permanganate etc.

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  4

(c)

Interpretation Introduction

Interpretation:

Structural formula of a compound with the given formula upon oxidation by potassium dichromate in aqueous sulphuric acid gives carboxylic acid has to be drawn.

Concept-Introduction:

Oxidation:

Loss of electrons from an atom ion or molecule during a chemical reaction is known as oxidation.  Oxidation state of atom ion or molecule will increase in this process.  Oxidizing agent is getting reduced in oxidation.  In simple it is the addition of oxygen or removal of hydrogen.

Example:

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  5

Here H2 oxidized to 2H+ by the removal of two electron.

Oxidation of alcohols:

Oxidation of alcohol gives aldehydes ketones carboxylic acid Primary alcohols on oxidation yield carboxylic acid.  Secondary alcohol on oxidation gives ketone.  In the formation of carboxylic acid the alcohol is first oxidized to aldehyde which is then oxidized to form carboxylic acid.  Oxidation is not possible in tertiary alcohol because there is no reactive C-H bond present adjacent to the C-OH group so oxidation will not be possible without breaking the bond.  Common oxidizing agents are pyridinium chlorochromate, dichromate, collins reagent, dimethyl sulfoxide, potassium permanganate etc.

Organic Chemistry, Chapter 17, Problem 17.21P , additional homework tip  6

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Chapter 17 Solutions

Organic Chemistry

Ch. 17 - Write the IUPAC name of each compound, showing...Ch. 17 - Prob. 17.8PCh. 17 - Prob. 17.9PCh. 17 - Prob. 17.10PCh. 17 - Prob. 17.11PCh. 17 - Prob. 17.12PCh. 17 - Prob. 17.13PCh. 17 - On a cyclohexane ring, an axial carboxyl group has...Ch. 17 - Prob. 17.15PCh. 17 - Prob. 17.16PCh. 17 - Prob. 17.17PCh. 17 - Complete each reaction.Ch. 17 - Prob. 17.19PCh. 17 - Prob. 17.20PCh. 17 - Prob. 17.21PCh. 17 - Show the reagents and experimental conditions...Ch. 17 - Prob. 17.23PCh. 17 - Prob. 17.24PCh. 17 - Prob. 17.25PCh. 17 - In each set, assign the acid its appropriate pKa.Ch. 17 - Low-molecular-weight dicarboxylic acids normally...Ch. 17 - Complete the following acid-base reactions. (a)...Ch. 17 - Prob. 17.29PCh. 17 - Prob. 17.30PCh. 17 - Excess ascorbic acid is excreted in the urine, the...Ch. 17 - Give the expected organic product when...Ch. 17 - Show how to convert trans-3-phenyl-2-propenoic...Ch. 17 - Show how to convert 3-oxobutanoic acid...Ch. 17 - Prob. 17.35PCh. 17 - Prob. 17.36PCh. 17 - Prob. 17.37PCh. 17 - When 4-hydroxybutanoic acid is treated with an...Ch. 17 - Fischer esterification cannot be used to prepare...Ch. 17 - Draw the product formed on thermal decarboxylation...Ch. 17 - Prob. 17.41PCh. 17 - Show how cyclohexanecarboxylic acid could be...Ch. 17 - Prob. 17.43PCh. 17 - Prob. 17.44PCh. 17 - Prob. 17.45PCh. 17 - Write the products of the following sequences of...Ch. 17 - Using your reaction roadmaps as a guide, show how...Ch. 17 - Using your reaction roadmaps as a guide, show how...Ch. 17 - Using your reaction roadmaps as a guide, show how...Ch. 17 - Using your reaction roadmaps as a guide, show how...Ch. 17 - Prob. 17.51PCh. 17 - Complete the following Fischer esterification...Ch. 17 - Prob. 17.53P
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